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Math Help - Decay Problem

  1. #1
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    Decay Problem

    "The half-life of radioactive carbon (14) is 5730 years. If the amount of radioactive carbon left over after 1000 years is 5 grams, how much radioactive carbon did you start with?"

    equation given to me is A = A_0e^{-kt}

    Now how do I find A_0

    I've found k = -0.000121

    Therefore I have formulated this equation

    5 = A_0e^{(-0.000121*1000)}

    My problem is, how do I find A_0 from the above formula or have I done something wrong. Everytime I do it, I end up with weird answer. Please help

    Just out of curiosity though, my answer = 5.643124562. Is that right? or wrong?
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  2. #2
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    you know how to solve that, as you did since you got a correct answer
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  3. #3
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    Quote Originally Posted by dwat View Post
    "The half-life of radioactive carbon (14) is 5730 years. If the amount of radioactive carbon left over after 1000 years is 5 grams, how much radioactive carbon did you start with?"

    equation given to me is A = A_0e^{-kt}

    Now how do I find A_0

    I've found k = -0.000121 Mr F says: There should be no negative. Clearly a typo on your part. And you should state the accuracy that you've rounded your answer to.

    Therefore I have formulated this equation

    5 = A_0e^{(-0.000121*1000)}

    My problem is, how do I find A_0 from the above formula or have I done something wrong. Everytime I do it, I end up with weird answer. Please help

    Just out of curiosity though, my answer = 5.643124562. Is that right? or wrong?
    5 = A_0 e^{-(0.000121)(1000)} = A_0 e^{-0.121} \Rightarrow A_0 = \frac{5}{e^{-0.121}} = 5.64 gram, correct to three significant figures.

    Since you have rounded your calculated value of k to three significant figures, your final answer is calculator madness - you cannot have an answer that is more accurate than the least accurate data used to get the answer ....
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    5 = A_0 e^{-(0.000121)(1000)} = A_0 e^{-0.121} \Rightarrow A_0 = \frac{5}{e^{-0.121}} = 5.64 gram, correct to three significant figures.

    Since you have rounded your calculated value of k to three significant figures, your final answer is calculator madness - you cannot have an answer that is more accurate than the least accurate data used to get the answer ....
    So I did get it right

    Oh sorry, yeah I meant k = 0.000121

    Well it asked to round k to 6 decimal places so I did. But you have a point. My mistake
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